1. III. Completely Randomized Design (CRD)
III.A Design of a CRD
III.B Models and estimation for a CRD
III.C Hypothesis testing using the ANOVA method
III.D Diagnostic checking
III.E Treatment differences
Statistical Modelling Chapter III

2. Diet A, B, C will have 3, 2, 1 rats, respectively.
III.A Design of a CRD
Definition III.1: An experiment is set up using a CRD when each treatment is applied a specified, possibly unequal, number of times, the particular units to receive a treatment being selected completely at random.
Example III.1 Rat experiment
Experiment to investigate 3 rat diets with 6 rats:
Diet A, B, C will have 3, 2, 1 rats, respectively.
Statistical Modelling Chapter III

3. Use R to obtain randomized layouts
How to do this is described in Appendix B, Randomized layouts and sample size computations in , for all the designs that will be covered in this course, and more besides.
Statistical Modelling Chapter III

4. R functions and output to produce randomized layout
> # Obtaining randomized layout for a CRD
> #
> n <- 6
> CRDRat.unit <- list(Rat = n)
> Diet <- factor(rep(c("A","B","C"),
> times = c(3,2,1)))
> CRDRat.lay <- fac.layout(unrandomized=CRDRat.unit,
> randomized=Diet, seed=695)
> CRDRat.lay
fac.layout from dae package produces the randomized layout.
unrandomized gives the single unrandomized factor indexing the units in the experiment.
randomized specifies the factor, Diets, that is to be randomized.
seed is used so that the same randomized layout for a particular experiment can be generated at a later date. (0–1023)
Statistical Modelling Chapter III

5. Randomized layout Units Permutation Rat Diet 1 1 4 1 A 2 2 1 2 CB A A B
#remove Diet object in workspace to avoid using it by mistake
remove(Diet)
Statistical Modelling Chapter III

6. III.B Models and estimation for a CRD
The analysis of CRD experiments uses:
least-squares or maximum likelihood estimation of the parameters of a linear model
hypothesis testing based on the ANOVA method or maximum likelihood ratio testing.
Use rat experiment to investigate linear models and the estimation of its parameters.
Statistical Modelling Chapter III

7. a) Maximal model
Definition III.2: The maximal expectation model is the most complicated model for the expectation that is to be considered in analysing an experiment.
We first consider the maximal model for the CRD.
Statistical Modelling Chapter III

8. Example III.1 Rat experiment (continued)
Suppose, the experimenter measured the liver weight as a percentage of total body weight at the end of the experiment.
The results of the experiment are as follows:
The analysis based on a linear model, that is:
Our model also involves assuming
Trick is what are X and q going to be?
Statistical Modelling Chapter III

9. Perhaps?
Note numbering of Y's does not correspond to Rats; does not affect model but neater.
This model can then be fitted using simple linear regression techniques.
The fitted equation is:
Statistical Modelling Chapter III

10. Using model to predict However, does this make sense?
means that, for each unit increase in diet, % liver weight decreases by
sensible only if the diets differences are based on equally spaced levels of some component;
For example, if the diets represent 2, 4 and 6 mg of copper added to each 100g of food
But, no good if diets unequally spaced (2, 4, and 10 mg Cu added) or diets differ qualitatively.
Statistical Modelling Chapter III

11. Regression on indicator variables
In this method the explanatory variables are called factors and the possible values they take levels.
Thus, we have a factor Diet with 3 levels: A, B, C.
Definition III.3: Indicator variables are formed from a factor:
create a variable for each level of the factor;
the values of a variable are either 1 or 0, 1 when the unit has the particular level for that variable and 0 otherwise.
Statistical Modelling Chapter III

12. Indicator-variable model
Hence
E[Yi] = ak, var[Yi] = s2, cov[Yi, Yj] = 0, (i  j)
Can be written as
Model suggests 3 different expected (or mean) values for the diets.
Statistical Modelling Chapter III

13. General form of X for CRD
For the general case of a set of t Treatments suppose Y is ordered so that all observations for:
1st treatment occur in the first r1 rows,
2nd treatment occur in the next r2 rows,
and so on with the last treatment occurring in the last rt rows.
i.e. order of systematic layout (prior to randomization)
Then XT given by the following partitioned matrix
(1 only ever vector, but 0 can be matrix)
Statistical Modelling Chapter III

14. Still a linear model
In general, the model for the expected values is still of the general form E[Y] = Xq
and on assuming Y is
can use standard least squares or maximum likelihood estimation
Statistical Modelling Chapter III

15. Estimates of expectation parameters
OLS equation is
Can be shown, by examining the OLS equation, that the estimates of the elements of a and y are the means of the treatments.
Example III.1 Rat experiment (continued)
The estimates of a are:
Statistical Modelling Chapter III

16. Example III.1 Rat experiment (continued)
The estimates of the expected values, the fitted values, are given by:
Statistical Modelling Chapter III

17. Estimator of the expected values
In general, where is the n-vector consisting of the treatment means for each unit and
being least squares, this estimator can be written as a linear combination of Y.
that is, can be obtained as the product of an matrix and the n-vector Y.
let us write
M for mean because MT is the matrix that replaces each value of Y with the mean of the corresponding treatment.
Statistical Modelling Chapter III

18. General form of mean operator
Can be shown that the general form of MT is
Clear from the above expression that:
1st r1 elements of are the mean of the Yis for the 1st treatment,
next r2 elements are the mean of those for the 2nd treatment,
and so on.
MT is a mean operator as it
computes the treatment means from the vector to which it is applied and
replaces each element of this vector with its treatment mean.
Statistical Modelling Chapter III

19. Estimator of the errors
The estimator of the random errors in the observed values of Y is, as before, the difference from the expected values.
That is,
Statistical Modelling Chapter III

20. Example III.1 Rat experiment Alternative expression for fitted values
We know that where
Note not as estimates rather than estimators.
Statistical Modelling Chapter III

21. Residuals
Fitted values for orthogonal experiments are functions of means.
Residuals are differences between observations and fitted values
Statistical Modelling Chapter III

22. b) Alternative indicator-variable, expectation models
For the CRD, two expectation models are considered:
First model is minimal expectation model: population mean response is same for all observations, irrespective of diet.
Second model is the maximal expectation model.
Statistical Modelling Chapter III

23. Minimal expectation model
Definition III.4: The minimal expectation model is the simplest model for the expectation that is to be considered in analysing an experiment.
The minimal expectation model is the same as the intercept-only model given for the single sample in chapter I, Statistical inference.
Will be this for all analyses we consider.
Now the estimator of the expected values in the intercept-only model is where is the n-vector each of whose elements is the grand mean.
For Rat experiment
Statistical Modelling Chapter III

24. Submodels
In regression case submodels have some of the parameters in the full model set to zero.
Here this is not the case.
Instead have that submodels are marginal to full models.
Here simply set ak = m
That is, intercept only model is the special case where all ak s are equal.
Clear for getting E[Yi] = m from E[Yi] = ak.
What about yG from yT?
If replace each element of a with m, then a = 1tm.
So yT = XTa = XT1tm = XGm.
Now marginality derives from the relationship between XT and XG as encapsulated in definition.
Statistical Modelling Chapter III

25. Marginality of models (in general)
Definition III.5: Let C(X) denote the column space of X.
For two models, y1  X1q1 and y2  X2q2, the first model is marginal to the second if C(X1)  C(X2) irrespective of the replication of the levels in the columns of the Xs,
That is if the columns of X1 can always be written as linear combinations of the columns of X2.
We write y1  y2.
Note marginality relationship is not symmetric — it is directional, like the less-than relation.
So while y1  y2, y2 is not marginal to y1 unless y1  y2.
Statistical Modelling Chapter III

26. Marginality of models for CRD
yG is marginal to yT or yG  yT because C(XG)  C(XT)
in that an element from a row of XG is the sum of the elements in the corresponding row of XT
and this will occur irrespective of the replication of the levels in the columns of XG and XT.
So while yG  yT, yT is not marginal to yG as C(XT)  C(XG) so that yT  yG.
In geometrical terms, C(XT) is a three-dimensional space and C(XG) is a line, the equiangular line, that is a subspace of C(XT).
Statistical Modelling Chapter III

27. III.C Hypothesis testing using the ANOVA method
Are there significant differences between the treatment means?
This is equivalent to deciding which of our two expectation models best describes the data.
We now perform the hypothesis test to do this for the example.
Statistical Modelling Chapter III

28. Analysis of the rat example Example III.1 Rat experiment (continued)
Step 1: Set up hypotheses
H0: aA = aB = aC = m (yG  XGm)
H1: not all population Diet means are equal
(yD  XDa)
Set a = 0.05
Statistical Modelling Chapter III

29. Example III.1 Rat experiment (continued)
Step 2: Calculate test statistic
From table can see that (corrected) total variation amongst the 6 Rats is partitioned into 2 parts:
variance of difference between diet means and
the left-over (residual) rat variation.
Step 3: Decide between hypotheses
As probability of exceeding F of 3.60 with n1 = 2 and n2 = 3 is > a = 0.05, not much evidence of a diet difference.
Expectation model that appears to provide an adequate description of the data is yG  XGm.
Statistical Modelling Chapter III

30. b) Sums of squares for the analysis of variance
From chapter I, Statistical inference, an SSq
is the SSq of the elements of a vector and
can write as the product of transpose of a column vector with original column vector.
Estimators of SSqs for the CRD ANOVA are SSqs of following vectors (cf ch.I):
where Ds are n-vectors of deviations from Y and
Te is the n-vector of Treatment effects.
Definition III.6: An effect is a linear combination of means with a set of effects summing to zero.
Statistical Modelling Chapter III

31. SSqs as quadratic forms
Want to show estimators of all SSqs can be written as YQY.
Is product of 1n, nn and n1 vectors and matrix, so is 11 or a scalar.
Definition III.7: A quadratic form in a vector Y is a scalar function of Y of the form YAY where A is called the matrix of the quadratic form.
Statistical Modelling Chapter III

32. SSqs as quadratic forms (continued)
Firstly write
That is, each of the individual vectors on which the sums of squares are based can be written as an M matrix times Y.
These M matrices are mean operators that are symmetric and idempotent: M' = M and M2 = M in all cases.
Statistical Modelling Chapter III

33. SSqs as quadratic forms (continued)
Then
Given Ms are symmetric and idempotent, it is relatively straightforward to show so are the three Q matrices.
It can also be shown that
Statistical Modelling Chapter III

34. SSqs as quadratic forms (continued)
Consequently obtain the following expressions for the SSqs:
Statistical Modelling Chapter III

35. SSqs as quadratic forms (continued)
Theorem III.1: For a completely randomized design, the sums of squares in the analysis of variance for Units, Treatments and Residual are given by the quadratic form:
Proof: follows the argument given above.
Statistical Modelling Chapter III

36. Residual SSq by difference
In the notes show that
so that
That is, Residual SSq = Units SSq - Treatments SSq.
Statistical Modelling Chapter III

37. ANOVA table construction
As in regression, Qs are orthogonal projection matrices.
QU orthogonally projects the data vector into the n-1 dimensional part of the n-dimensional data space that is orthogonal to equiangular line.
QT orthogonally projects data vector into the t-1 dimensional part of the t-dimensional Treatment space, that is orthogonal to equiangular line. (Here the Treatment space is the column space of XT.)
Finally, the matrix orthogonally projects the data vector into the n-t dimensional Residual subspace.
That is, Units space is divided into the two orthogonal subspaces, the Treatments and Residual subspaces.
Statistical Modelling Chapter III

38. Geometric interpretation
Of course, the SSqs are just the squared lengths of these vectors
Hence, according to Pythagoras’ theorem, the Treatments and Residual SSqs must sum to the Units SSq.
Statistical Modelling Chapter III

39. Example III.1 Rat experiment (continued)
Vectors for computing the SSqs are:
Total Rat deviations, Diet Effects and Residual Rats deviations are projections into Rats, Diets and Residual subspaces of dimension 5, 2 and 3, respectively.
Squared length of projection = SSq
Rats SSq is Y'QRY = 0.34
Diets SSq is Y'QDY = 0.24
Residual SSq is
Exercise III.3 is similar example for you to try
Statistical Modelling Chapter III

40. c) Expected mean squares
Have an ANOVA in which we use F (= ratio of MSqs) to decide between models.
But why is this ratio appropriate?
One way of answering this question is to look at what the MSqs measure?
Use expected values of the MSqs, i.e. E[MSq]s, to do this
Statistical Modelling Chapter III

41. Expected mean squares (cont’d)
Need E[MSq]s
under the maximal model:
and the minimal model:
Similar to asking what is E[Yi]?
Know answer is E[Yi] = ak.
i.e. in population, under model, average value of Yi is ak.
So for Treatments, what is E[MSq]?
The E[MSq]s are the mean values of the MSqs in populations described by the model for which they are derived
i.e. an E[MSq] is the true mean value;
it depends on the model parameters.
Statistical Modelling Chapter III

42. Expected mean squares (cont’d)
Need E[MSq]s
under the maximal model:
and the minimal model:
The E[MSq]s are the mean values of the MSqs in populations described by the model for which they are derived
i.e. an E[MSq] is the true mean value and it depends on the model parameters.
This is analogous to saying that the expected value of the Treatment mean vector in populations described by the model is y.
That is,
Statistical Modelling Chapter III

43. Expected mean squares (cont’d)
Need E[MSq]s
under the maximal model:
and the minimal model:
The E[MSq]s are the mean values of the MSqs in populations described by the model for which they are derived
This is analogous to saying that the expected value of the Treatment mean vector in populations described by the model is y.
That is,
Put another way, what is the mean of the sampling distribution of:
the Treatment mean vector?
the Treatment mean square?
(Note that both the mean vector and the mean square are functions of Y and so are random variables.)
The answer is easy for the treatment mean vector: y
But what for the Treatment mean square? Ans. E[MSq]
Statistical Modelling Chapter III

44. E[MSq]s under the maximal model
So if we had the complete populations for all Treatments and computed the MSqs, the value of
the Residual MSq would equal s2
the Treatment MSq would equal s2 + qT(y).
So that the population average value of both MSqs involves s2, the uncontrolled variation amongst units from the same treatment.
But what about q in Treatments E[MSq].
Statistical Modelling Chapter III

45. The qT(y) function
Subscript T indicates the Q matrix on which function is based:
but no subscript on the y in qT(y),
because we will determine expressions for it under both the maximal (yT) and alternative models (yG).
That is, y in qT(y) will vary.
Numerator is same as the SSq except that it is a quadratic form in y instead of Y.
To see what this means want expressions in terms of individual parameters.
Will show that under the maximal model (yT)
and under the minimal model (yG) that
Statistical Modelling Chapter III

46. Example III.1 Rat experiment (continued)
The latter is just the mean of the elements of yT.
Actually, the quadratic form is the SSQ of the elements of vector
When will the SSq be zero?
Statistical Modelling Chapter III

47. The qT(y) function qT(y) is the SSq of QTy, divided by (t-1).
Now want to prove the following result:
As QT is symmetric and idempotent,
y'QTy = (QTy)'QTy
qT(y) is the SSq of QTy, divided by (t-1).
QTy = (MT – MG)y = MTy – MGy
MGy replaces each element of y with the grand mean of the elements of y
MTy replaces each element of y with the mean of the elements of y that received the same treatment as the element being replaced.
Statistical Modelling Chapter III

48. The qT(y) function (continued)
Under the maximal model (yT)
so that y'TQTyT is the SSq of the elements of
Under the minimal model (yG = m1n)
MGyG = MTyG = yG so y'GQTyG = 0 and qT(y) = 0;
or ak = m so that
and so that qT(y) = 0.
Statistical Modelling Chapter III

49. Example III.1 Rat experiment (continued)
Statistical Modelling Chapter III

50. How qT(yT) depends on the as
qT(yT) is a quadratic form and is basically a sum of squares so that it must be nonnegative.
Indeed the magnitude of depends on the size of the differences between the population treatment means, the aks
if all the aks are similar they will be close to their mean,
whereas if they are widely scattered several will be some distance from their mean.
Statistical Modelling Chapter III

51. E[Msq]s in terms of parameters
Could compute population mean of MSq if knew aks and s2.
Treatment MSq will on average be greater than the Residual MSq
as it is influenced by both uncontrolled variation and the magnitude of treatment differences.
The quadratic form qT(y) will only be zero when all the as are equal, that is when the null hypothesis is true.
Then the E[MSq]s under the minimal model are equal so that the F value will be approximately one.
Not surprising if think about a particular experiment.
Statistical Modelling Chapter III

52. Example III.1 Rat experiment (continued)
So what can potentially contribute to the difference in the observed means of 3.1 and 2.7 for diets A and C?
Answer:
Obviously, the different diets;
not so obvious that differences arising from uncontrolled variation also contribute as 2 different groups of rats involved.
This is then reflected in E[MSq] in that it involves s2 and the "variance" of the 3 effects.
Statistical Modelling Chapter III

53. Justification of F-test
Thus, the F test involves asking the question "Is the variance in the sample treatment means > can be expected from uncontrolled variation alone?".
If the variance is no greater, concluded qT(y) = 0
the minimal model is the correct model since the expected Treatment MSq under this model is just s2.
Otherwise, if the variance is greater, qT(y) is nonzero
the maximal model is required to describe the data.
Similar argument to examining dotplots for Example II.2, Paper bag experiment.
Statistical Modelling Chapter III

54. d) Summary of the hypothesis test
see notes
Statistical Modelling Chapter III

55. d) Summary of the hypothesis test
Summarize the ANOVA-based hypothesis test for a CRD involving t treatments and a total of n observed units.
Step 1: Set up hypotheses
H0: a1 = a2 =...= a3 = m (yG  XGm)
H1: not all population Treatment means are equal (yT  XTa)
Set a.
Statistical Modelling Chapter III

56. Summary (continued) Step 2: Calculate test statistic
Note that E[MSq]s under the maximal model are included in this table, it being recognized that when H0 is true qT(y) = 0.
Step 3: Decide between hypotheses
Determine probability of observed F value that has n1 = numerator d.f. = t‑1 and n2 = denominator d.f. = n-t. If
then the evidence suggests that the null hypothesis be rejected and the alternative model be used to describe the data.
Statistical Modelling Chapter III

57. e) Comparison with traditional one-way ANOVA
Our ANOVA table is essentially the same as the traditional table
the values of the F statistic from each table are exactly the same.
Labelling differs and the Total would normally be placed at the bottom of the table, not at the top.
Difference is symbolic:
Units term explicitly represents a source of uncontrolled variation: differences between Units.
Our table exhibits the confounding in the experiment.
Indenting of Treatments under Units signifies that treatment differences are confounded or “mixed-up” with unit differences.
Statistical Modelling Chapter III

58. f) Computation of the ANOVA in R
Begin with data entry (see Appendix A, Introduction to R and, Appendix C, Analysis of designed experiments in R)
Next an initial graphical exploration using boxplots — defer to a second example with more data.
ANOVA: while function lm could be used, function aov is preferred for analysing data from a designed experiment.
Both use a model formula of the form:
Response variable ~ explanatory variables (and operators)
So far expressions on right fairly simple — one or two explanatory variables separated by a “+”.
Subtlety with analysis of designed experiments:
If explanatory variable is a numeric, such as a numeric vector, then R fits just one coefficient for it.
For a single explanatory variable, a straight-line relationship fitted.
If explanatory variable is categorical, such as a factor, a coefficient is fit for each level of the variable — indicator variables are used.
In analyzing a CRD important that the treatment factor is stored in a factor object, signalling R to use indicator variables
Statistical Modelling Chapter III

59. f) Computation of the ANOVA in R (continued)
There are two ways in which this analysis can be obtained using the aov function:
without and with an Error function in the model formula.
Error function used in model formula to specify a model for the Error in the experiment (a model for uncontrolled variation).
summary function is used and this produces ANOVA table.
model.tables function used to obtain tables of means.
Statistical Modelling Chapter III

60. Example III.1 Rat experiment (continued)
Following commands to perform the two analyses of the data: #
# AOV without Error
Rat.NoError.aov <- aov(LiverWt ~ Diet, CRDRat.dat)
summary(Rat.NoError.aov)
# AOV with Error
Rat.aov <- aov(LiverWt ~ Diet + Error(Rat), CRDRat.dat)
summary(Rat.aov)
model.tables(Rat.aov, type = "means")
Statistical Modelling Chapter III

61. Output
> # AOV without Error
> #
> Rat.NoError.aov <- aov(LiverWt ~ Diet, CRDRat.dat)
> summary(Rat.NoError.aov)
Df Sum Sq Mean Sq F value Pr(>F)
Diet
Residuals
> # AOV with Error
> Rat.aov <- aov(LiverWt ~ Diet + Error(Rat), CRDRat.dat)
> summary(Rat.aov)
Error: Rat
Diet
Residuals
Analysis without the Error function parallels the traditional analysis and that with Error is similar to our table.
In this course will use Error function.
Statistical Modelling Chapter III

62. Output (continued) > model.tables(Rat.aov, type="means")
Tables of means
Grand mean
3.1
Diet
A B C
rep
Statistical Modelling Chapter III

63. III.D Diagnostic checking
Assumed the following model:
Y is distributed N(y, s2In)
where y = E[Y] = Xq and
s2 is the variance of an observation
Maximal model is used in diagnostic checking:
yT = E[Y] = XTa
For this model to be appropriate requires that:
the response is operating additively — that the individual deviations in the response to a treatment are similar;
the sets of units assigned the treatments are comparable in that the amount of uncontrolled variation exhibited by them is the same for each treatment;
each observation is independent of other observations; and
that the response of the units is normally distributed.
Statistical Modelling Chapter III

64. Example III.2 Caffeine effects on students
Effect of orally ingested caffeine on a physical task was investigated (Draper and Smith, 1981, sec.9.1).
Thirty healthy male college students were selected and trained in finger tapping.
Ten men were randomly assigned to receive one of three doses of caffeine (0, 100 or 200 mg).
The number of finger taps after ingesting the caffeine was recorded for each student.
Statistical Modelling Chapter III

65. Entering the data
Setting up a data frame for data arranged in standard order:
factor Students with values 1–30,
factor Dose with levels 0, 100 and 200 and values depending on whether data is entered by rows or columns (can use rep function),
numeric vector Taps with the 30 observed values of the response variable.
Statistical Modelling Chapter III

66. Entering the data (cont'd)
> CRDCaff.dat
Students Dose Taps
Entering the data (cont'd)
#set up data.frame with factors Students and Dose and response variable Taps
CRDCaff.dat <- data.frame(Students = factor(1:30),
Dose = factor(rep(c(0,100,200), times=10)))
CRDCaff.dat$Taps <-
c(242,248,246,245,246,248,244,245,250,248,247,252,247,248,248,
248,250,250,242,247,246,244,246,248,246,243,245,242,244,250)
CRDCaff.dat
Statistical Modelling Chapter III

67. Boxplots for each level of Dose
Use function
boxplot(split(Taps, Dose), xlab="Dose", ylab="Number of taps")
Average number of taps increasing as dose increases
Some evidence dose 0 more variable than dose 100.
Statistical Modelling Chapter III

68. Analysis of variance for this data
> Caffeine.aov <- aov(Taps ~ Dose + Error(Students), CRDCaff.dat)
> summary(Caffeine.aov)
Error: Students
Df Sum Sq Mean Sq F value Pr(>F)
Dose
Residuals
> model.tables(Caffeine.aov, type="means")
Tables of means
Grand mean
246.5
Dose
Statistical Modelling Chapter III

69. The hypothesis test Step 1: Set up hypotheses
H0: a0 = a100 = a200 = m (yG = XGm)
H1: not all population Dose means are equal
(yD = XDa)
Set a = 0.05
Statistical Modelling Chapter III

70. The hypothesis test (continued)
Step 2: Calculate test statistic
  The ANOVA table for the example is:
Step 3: Decide between hypotheses
P(F2,27  6.18) = p = < a = 0.05.
  The evidence suggests there is a dose difference and that the expectation model that best describes the data is yD  XDa.
Statistical Modelling Chapter III

71. Examination of the residuals, eT
Use the Residuals-versus-fitted-values plot and the Normal Probability plot.
In interpreting these plots:
Note ANOVA is robust to variance heterogeneity, if treatments equally replicated, and to moderate departures from normality.
Most commonly find an unusually large or small residual.
The cause of such extreme values requires investigation.
May be a recording mistake or catastrophe for a unit that can be identified.
But, may be valid and is the result of some unanticipated, but important effect.
Statistical Modelling Chapter III

72. The Normal Probability plot
Should show a broadly straight line trend. *
* *
__________________
Statistical Modelling Chapter III

73. The Residuals-versus-fitted-values plot
Generally, the points on scatter diagram should be spread across plot evenly.
  * * *
* * *
* *
* * *
* * *
_____________________________
no particular pattern
Statistical Modelling Chapter III

74. Problem plots * * * systematic trend in residuals
  * *
* * *
* * * *
* *
* *
* * *
_________________________
systematic trend in residuals *
* *
* *
* *
__________________________
variance increases as level increases *
* *
* * *
* * * * *
* *
_________________________
variance peaks in middle
Actually, for the CRD, this plot has a vertical scatter of points for each treatment — each should be centred on zero and of the same width.
Statistical Modelling Chapter III

75. R functions used in producing these plots
resid.errors: extract the residuals from an aov object when Error function used
fitted.errors: extract the fitted values from an aov object when Error function used
plot: to plot the fitted values against the residuals
qqnorm: to plot the residuals against the normal quantiles
qqline: to add a line to the plot produced by qqnorm.
First 2 functions are nonstandard functions from dae package.
Statistical Modelling Chapter III

76. Example III.2 Caffeine effects on students (continued)
A violation of the assumptions would occur if all the students were in the same room and the presence of other students caused anxiety to just the students that had no caffeine.
The response of the students is not independent.
It may be that the inhibition of this group resulted in less variation in their response which would be manifest in the plot.
Another situation that would lead to an unacceptable pattern in the plot is if the effect becomes more variable as the level of the response variable increases.
For example, caffeine increases the tapping but at higher levels the variability of increase from student to student is greater.
That is there is a lack of additivity in the response.
Statistical Modelling Chapter III

77. R output for getting the plots
> res <- resid.errors(Caffeine.aov)
> fit <- fitted.errors(Caffeine.aov)
> data.frame(Students,Dose,Taps,res,fit)
Students Dose Taps res fit
Note the use of data.frame to produce a printed list of the original data with the residuals and fitted values.
Statistical Modelling Chapter III

78. Plots for the example
> plot(fit, res, pch = 16)
> qqnorm(res, pch = 16)
> qqline(res)
The Residuals-versus-fitted-values plot appears to be fine.
Statistical Modelling Chapter III

79. Normal Probability plot
Displaying some curvature at ends.
Indicates data heavier in tails and flatter in the peak than expected for a normal distribution.
Given normality not crucial and only a few observations involved, use analysis we have performed.
Statistical Modelling Chapter III

80. The hypothesis test – Summary
The ANOVA table for the example is:
P(F2,27  6.18) = p = < a = 0.05.
  The evidence suggests there is a dose difference and that the expectation model that best describes the data is yD  XDa.
Diagnostic checking indicates model is OK
Statistical Modelling Chapter III

81. III.E Treatment differences
So far all that our analysis has accomplished is that we have decided whether or not there appears to be a difference between the population treatment means.
Of greater interest to the researcher is how the treatment means differ.
Two alternatives available:
Multiple comparisons procedures
used when the treatment factors are all qualitative so that it is appropriate to test for differences between treatments
Fitting submodels
When one (or more) of the treatment factors is quantitative the fitting of smooth curves to the trend in the means is likely to lead to a more appropriate and concise description of the effects of the factors.
Often, for reasons explained in chapter II, a low order polynomial will provide an adequate description of the trend.
Statistical Modelling Chapter III

82. Note
Multiple comparison procedures should not be used when the test for treatment differences is not significant.
Submodels should be fitted irrespective of whether the overall test for treatment differences is significant.
The difference in usage has to do with one being concerned with mean differences and the other with deciding between models.
Statistical Modelling Chapter III

83. a) Multiple comparisons procedures for comparing all treatments
Multiple comparisons for all treatments = MCA procedures.
In general MCA procedures divide into those based:
on family-wise error rates — Type I error rate specified and controlled for over all comparisons, often at 0.05.
those based on comparison-wise error rates — Type I error rate specified and controlled for each comparison
Problem with latter is probability of an incorrect conclusion gets very high as the number of comparisons increases.
For comparison-wise error rate of 0.05, family-wise error rate:
So recommend use MCA procedure based on family-wise error rates: use just Tukey's HSD procedure.
Statistical Modelling Chapter III

84. Tukey’s Honestly Significant Difference procedure
determines if each pair of means is significantly different
is based on a family-wise error rate.
basically for equal numbers of observations for each mean.
A modification that is approximate will be provided for unequal numbers.
Statistical Modelling Chapter III

85. The procedure
Each application of the procedure is based on the hypotheses:
Ho: aA = aB
H1: aA  aB
One calculates the statistic:
qt,n,a is the studentized range with t = no. means, n = Residual df and a = significance level,
is the standard error of the difference
rA, rB are the number of replicates for each of a pair of means being compared.
Statistical Modelling Chapter III

86. Notes about replication
Note that strictly speaking rA and rB should be equal.
When unequal rA and rB called the Tukey-Kramer procedure.
w(a%) will depend on which means are being compared.
If the treatments are all equally replicated with replication r, the formula for reduces to
Statistical Modelling Chapter III

87. In R aov has given Residual MSq and df
model.tables produces tables of means.
qtukey computes as follows:
q <- qtukey(1 - a, t, n)
Statistical Modelling Chapter III

88. Example III.2 Caffeine effects on students (continued)
Have already concluded evidence suggests that there is a dose difference. But which doses are different?
Output with means and q:
> model.tables(Caffeine.aov, type = "means")
Tables of means
Grand mean
246.5
Dose
> q <- qtukey(0.95, 3, 27)
> q
[1]
Statistical Modelling Chapter III

89. Decide on differences
Any two means different by 2.47 or more are significantly different.
Our conclusion is that the mean for 0 and 200 are different but that for 100 is somewhat intermediate.
Statistical Modelling Chapter III

90. b) Fitting submodels
For quantitative factors, like Dose, it is often better to examine the relationship between the response and the levels of the factor.
This is commonly done using polynomials.
Now, a polynomial of degree t-1 will fit exactly t points.
So a quadratic will fit exactly the three means.
In practice, polynomials of order 2 often sufficient.
However, more than 3 points may be desirable so that deviations from the fitted curve or lack of fit can be tested
Statistical Modelling Chapter III

91. Polynomial models
To investigate polynomial models up to order 2, following models for expectation investigated:
In matrix terms
where
xk is the value of the kth level of the treatment factor,
m is the intercept of the fitted equation and
g1 is the slope of the fitted equation and
g2 is the quadratic coefficient of the fitted equation.
The X1 and X2 matrices made up of columns that consist of the values of the levels of the factor and their powers
— not the indicator variables of before.
Statistical Modelling Chapter III

92. Example III.2 Caffeine effects on students (continued)
The X matrices for the example are:
The columns of each X matrix in the above list are a linear combination of those of any of the X matrices to its right in the list.
Statistical Modelling Chapter III

93. Marginality of models Each model in the sequence of models
is marginal to all models before it as
That is, the columns of each X matrix in the above list are a linear combination of those of any of the X matrices to its right in the list.
Marginality is not a symmetric relationship in that
if a model is marginal to second model,
the second model is not necessarily marginal to the first.
For example, E[Y] = X1g1 is marginal to E[Y] = X2g2 but not vice-a-versa, except when t = 2.
Statistical Modelling Chapter III

94. Equivalence of E[Y] = X2g2 and E[Y] = XDa
C(X2) = C(XD) as the 3 columns of one matrix can be written as 3 linearly independent combinations of the columns of the other matrix.
So the two models are marginal to each other and are equivalent
However, while the fitted values are the same, the estimates and interpretation of the parameters are different:
those corresponding to X2 are interpreted as the intercept, slope and curvature coefficient.
those corresponding to XD are interpreted as the expected (mean) value for that treatment.
Also, in spite of being marginal, the estimators of the same parameter differ depending on the model that has been fitted.
for example, for the model E[Y] = XGm,
but for E[Y] = X1g1,
models not orthogonal
Statistical Modelling Chapter III

95. Hypothesis test incorporating submodels
Test statistics computed in ANOVA table.
Step 1: Set up hypotheses
  a) H0:
H1:
  (Differences between fitted model or Deviations from quadratic are zero)
b) H0:
H1:  
c) H0:
Set a.
Statistical Modelling Chapter III

96. Hypothesis test (continued)
Step 2: Calculate test statistics
The ANOVA table for a CRD is:
Test statistics corresponds to hypothesis pairs in Step 1.
Statistical Modelling Chapter III

97. Hypothesis test (continued)
Step 3: Decide between hypotheses
Begin with the first hypothesis pair, determine its significance and continue down the sequence until a significant result is obtained.
A significant Deviations F indicates that the linear and quadratic terms provide an inadequate description.
Thus a model based on them would be unsatisfactory. No point in continuing.
If the Deviations F is not significant , then a significant Quadratic F indicates that a 2nd degree polynomial is required to adequately describe the trend.
As a linear coefficient is necessarily incorporated in a 2nd polynomial, no point to further testing in this case.
If both the Deviations and Quadratic F's are not significant, then a significant Linear F indicates a linear relationship describes the trend in the treatment means.
Statistical Modelling Chapter III

98. Fitting polynomials in R
Need to realize that the default contrasts for ordered (factor) objects are polynomial contrasts, assuming equally-spaced levels:
linear transformations of all columns of X, except the first, such that each column
is orthogonal to all other columns in X and
has SSq equal to 1.
Facilitates the computations.
So, if a factor is quantitative:
set it up as an ordered object from the start.
if did not make it an ordered object, then redefine factor as an ordered.
Statistical Modelling Chapter III

99. Example III.2 Caffeine effects on students (continued)
R output for setting up ordered (factor):
> # fit polynomials
> #
> Dose.lev <- c(0,100,200)
> CRDCaff.dat$Dose <- ordered(CRDCaff.dat$Dose,
levels=Dose.lev)
> contrasts(CRDCaff.dat$Dose) <- contr.poly(t,
scores=Dose.lev)
> contrasts(CRDCaff.dat$Dose)
.L Q e e e
Dose levels stored to save repeatedly entering them
redundant for equally-, but not unequally-, spaced levels.
coded values of dose and dose squared with S = 0, SSq = 1, S cross-product= 0
Statistical Modelling Chapter III

100. R output for fitting polynomial submodels
> Caffeine.aov <- aov(Taps ~ Dose + Error(Students), CRDCaff.dat)
> summary(Caffeine.aov, split = list(Dose = list(L = 1, Q = 2)))
Error: Students
Df Sum Sq Mean Sq F value Pr(>F)
Dose
Dose: L
Dose: Q
Residuals
Note 3 treatments will follow a quadratic exactly — this is reflected in C(X2) = C(XD).
Thus, Deviations line is redundant in this example.
When required, Deviations line is computed by assigning extra powers to a single line named, say, Dev in split function (e.g. Dev = 3:6 — see notes).
Statistical Modelling Chapter III

101. Hypothesis test incorporating submodels
Step 1: Set up hypotheses
a) H0:
H1:  
b) H0:
Set a = 0.05.
Statistical Modelling Chapter III

102. Hypothesis test (continued)
Step 2: Calculate test statistics
> Caffeine.aov <- aov(Taps ~ Dose + Error(Students), CRDCaff.dat)
> summary(Caffeine.aov, split = list(Dose = list(L = 1, Q = 2)))
Error: Students
Df Sum Sq Mean Sq F value Pr(>F)
Dose
Dose: L
Dose: Q
Residuals
Statistical Modelling Chapter III

103. Hypothesis test (continued)
Step 3: Decide between hypotheses
  The Quadratic source has a probability of > 0.05 and so H0 is not rejected in this case.
  The linear source has a probability of < 0.05 and so H0 is rejected in this case.
  It is clear quadratic term is not significant but linear term is highly significant so the appropriate model for the expectation is the linear model y = X1g1 where g1 = [m g1].
Statistical Modelling Chapter III

104. Fitted equation
Coefficients can be obtained using the coef function on the aov object, but these are not suitable for obtaining the fitted values.
The fitted equation is obtained by putting the values of the levels into a numeric vector and using the lm function to fit a polynomial of the order indicated by the hypothesis test.
For the example, linear equation was adequate and so the analysis is redone with 1 for the order of the polynomial.
Statistical Modelling Chapter III

105. Fitted equation for the example
> D <- as.vector(Dose)
> D <- as.numeric(D)
> Caffeine.lm <- lm(Taps ~ D)
> coef(Caffeine.lm)
(Intercept) D
The fitted equation is Y = X where X is the number of taps
The slope of this equation is
That is, taps increase  x 100 = 1.75 with each 100 mg of caffeine.
This conclusion seems a more satisfactory summary of the results than that the response at 200 is significantly greater than at 0 with 100 being intermediate.
The commands to fit a quadratic would be:
D2 <- D*D
Caffeine.lm <- lm(Taps ~ D + D2)
Statistical Modelling Chapter III

106. Plotting means and fitted line
Details are in the notes
The plot produced is as follows:
Statistical Modelling Chapter III

107. c) Comparison of treatment parametrizations
Two alternative bases:
Lead to different parameter estimates with different interpretations and different partitions of the treatment SSqs.
The total treatment SSqs and fitted values for treatments remain the same, while the contrasts span the treatment space.
That is, in this case, SSqT = SSqL + SSqQ
Statistical Modelling Chapter III

108. III.G Exercises Ex. III.1–2 look at aspects of quadratic forms
Ex. III.3 investigates the calculations with example that can be done with a calculator
Ex. III.4 involves producing a randomized layout
Ex. III.5 asks for the complete analysis of a CRD with a qualitative treatment factor
Ex. III.6 asks for the complete analysis of a CRD with a quantitative treatment factor
Statistical Modelling Chapter III